jmc

First, we note $$x^3-y^3=(x-y)(x^2+xy+y^2)=6(24+xy),$$ so we just need to find $xy$ now. Squaring both sides of $x-y=6$ gives $$x^2-2xy+y^2=36.$$ Since $x^2+y^2=24$, we have $24-2xy=36$, so $xy=-6$, from which we have $$x^3-y^3=6(24+xy)=6(24-6)=6(18)=\boxed{108}.$$